homework helper, contact and send problems

homework by email, send
helper, math and physics
math tutor web site
physics help web site
math and physics help, fees
help contact
problems, send here
forward payments here
teacher background

Exercises Using Ohmís Law

Homework Exercises, Ohm's Law,
Example One

The circuit below is called a Wheatstone bridge. It is used for measuring resistance. Show that when the current through the galvanometer G is zero, then R1 = R2(R3/R4). Thus, if we know R2 and the ratio (R3/R4), we can obtain the resistance R1.

IG = 0 means points C and D are at the same potential and
that I1 = I2, I3 = I4.

By Ohmís law VAC = I1R1, VAD = I3R3, VCB = I2R2, VDB = I4R4.

As VAC = VAD and VCB = VDB, then I1R1= I3R3,
I2R2= I4R4.

As I1 = I2 and I3 = I4, then doing I1R1/ I2R2=I3R3/I4R4 we get R1= R2(R3/R4).

Homework Exercises, Ohm's Law,
Example Two

Determine the total resistance and potential difference between points a and b in the circuit shown. Also determine the current in, and the potential difference across, each resistor. The input current is 60 A.

a) Total resistance between a and b.

First, calculate the 12, 6, and 4 ohm parallel combination R//1
1/R//1 = 1/12 + 1/6 + 1/4 = 1/2. Then R//1 = 2 ohm.

The upper branch is 3 ohms in series with R//1, 3 ohm + 2 ohm = 5 ohm.

This 5 ohm resistor is in parallel with the 20 ohm resistor, equivalent to a single resistor R//2 = 5 ohm//20 ohm:
1/R//2 = 1/5 + 1/20 = ľ. Then R//2 = 4.
The total resistance between a and b is 4 ohm + 5 ohm = 9 ohms. The voltage drop or potential difference between points a and b is 60Ax9ohm = 540 V.

b) To obtain the individual currents and voltages in each resistor we can proceed as follows (there is no a unique method):

The voltage across R//2 is the difference between the total 540 voltage and the voltage drop in the 5 ohm resistor. This drop is 60Ax5ohm = 300 V. So the potential difference across R//2 is 540 V Ė 300 V = 240 V.
Now we can calculate the currents and voltages in the R//2 combination:
In the 20 ohm resistor: current= 240 V/20 ohm = 12 A, voltage = 240 V
The upper branch has then a current of 60 A - 12 A = 48 A, or alternatively 240 V/5 ohm = 48A.
The 3 ohm resistor has a voltage of 48 Ax3 ohm = 144 V.
The R//1 combination presents a voltage of 48 AxR//1 = 48 Ax2 ohm = 96 V, or alternatively 240 V - 144 V = 96 V.
The currents in the 12 ohm, 6 ohm and 4 ohm are respectively 96/12 A=8 A, 96/6=16 A and 96/4=24 A.

Homework Exercises, Ohm's Law,
Example Three

Given the resistor arrangement shown below, prove that the relation between R1 and R2 must be R2=1.618R1 in order that the resistance of the system R be equal to R2.

The condition is R=R2, so we do
0= R22 - R1R2 - R12.

This is a second degree equation in R2, whose solution is R2= R1(1/2 Ī )

Disregarding the minus sign we get R2 = 1.618R1.

Ohm's Law Principle

Related Sites:
· Physics, Main Page
· Physics, Mathematics
· Physics, Detailed Homework Scope Help
· Energy, Work and Power: Concepts
· Kinetic Energy
· Potential Energy
· Power
· Physics Problems, Example
· Physics Homework - Mechanical Energy Conservation Problems
· Physics Homework - Mechanical Power Problems
· Coulomb's Law
· Exercises Using Coulomb's Law
· Electric Field Charges
· Electric Field Exercises
· Electric Potential Energy
· Exercises, Electric Potential Energy
· Gauss' Law
· Gauss' Law Exercises
· Second Newton's Law
· Second Newton's Law Examples, Part One
· Second Newton's Law Examples, Part Two
· Sound Waves
· Sound Waves: Standing, Interference, Doppler Effect - Examples
· Sound Waves, Doppler Effect - Examples
· Vectors, Scalars
· Vectors, Scalars - Analytic Method
· Addition Vector Tools, Problems
· Free Fall Theory
· Free Fall Exercises, Part One
· Free Fall Exercises, Part Two
· Free Fall Exercises, Part Three

Bookmark and Share